Re: origin of inertia
From: Dirk Van de moortel (dirkvandemoortel_at_ThankS-NO-SperM.hotmail.com)
Date: 03/14/05
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Date: Mon, 14 Mar 2005 14:49:40 GMT
<aleksandar.vukelja@gmail.com> wrote in message news:1110803443.295553.110290@o13g2000cwo.googlegroups.com...
> New theory introduces geometrical model of field, which makes inertia
> an obvious and imminent property of matter. Some related issues, such
> as relativity and gravity are also discussed, with some original
> conclusions.
>
> Visit http://www.masstheory.org and download pdf of the first chapter.
> The work has been accepted by The general Science Journal
> (http://www.wbabin.net) and will be discussed at NPA (Natural
> Philosophy Alliance) conference in May 2005.
Extract from section 2 of
http://www.masstheory.org/oos_1_english.pdf
| "So we get the following connection between coordinates:
| Dx' = 1/2 Dx
| Dt' = 2 Dt
| Now what does it mean? According to the theory, it means
| exactly the following: space in coordinate system of
| person B (marked with single quote), as observed and
| measured by person A, contracted by half to accommodate
| requirement of theory that speed of light remains constant.
| The time in coordinate system of person B, observed and
| measured by person A slowed to accommodate requirement
| of theory that speed of light remains constant."
This is very wrong. It is a very common mistake.
According to special relativity, the equations:
Dx' = 1/2 Dx
Dt' = 2 Dt
can only be valid together for the trivial case where everything
is zero:
Dx' = Dt' = Dx = Dt = 0.
Proof:
We take any pair of events E1 and E2 in spacetime.
Event E1 has coordinates (x1,t1) in the unprimed frame
and (x1',t1') in the primed frame.
Event E2 has coordinates (x2,t2) in the unprimed frame
and (x2',t2') in the primed frame.
Then we have the standard Lorentz transformation:
Dx' = g ( Dx - v Dt ) [1]
Dt' = g ( Dt - v Dx / c^2 ) [2]
with g = 1/sqrt(1-v^2/c^2) [3]
Dx = g ( Dx' + v Dt' ) [4]
Dt = g ( Dt' + v Dx' / c^2 ) [5]
where we have used the abbreviations:
Dx = x2-x1 Dt = t2-t1
Dx' = x2'-x1' Dt' = t2'-t1'
Suppose that v <> 0, otherwise we have only one frame.
You write
Dx' = Dx sqrt(1-v^2/c^2) [A]
which comes from
Dx' = 1/g Dx
i.o.w.
Dx = g Dx' ,
which implies through equation [4] that
Dt' = 0 [6]
This means that [A] is valid when the length of the rod
that is at rest in the unprimed frame, is measured by
taking two points simultaneously in the primed frame.
You also write
Dt' = Dt / sqrt(1-v^2/c^2) [B]
which comes from
Dt' = g Dt ,
which implies through equation [2] that
Dx = 0 [7]
This means that the ticks of the clock are measured
at the same place in the unprimed frame, i.o.w. the
clock is at rest in the unprimed frame.
So, when you take equations [A] and [B] together,
you also must bring equations [6] and [7] along, and
you necessarily end up two events that must satisfy
Dx' = Dt' = Dx = Dt = 0,
So, your equations [A] and [B], when considered together
amount to
0 = 0 [A]
0 = 0 [B]
which doesn't tell us much.
In other words still [A] and [B] are only valid for two
events E1 and E2 if
E1 = E2,
i.o.w. they cannot be valid for two distinct events.
I repeat, you made a very common mistake.
It is always caused by not understanding what the coordinates
(let alone the transformations) mean.
Most people who make the mistake, will build on it, and
quickly arrive at contradictions, which is what you do in
theorem 2.1.
This is very typical for the NPA. They think they fight relativity,
but, with or without knowing, they actually fight their own sad
(or deliberate) misunderstanding of it.
Dirk Vdm
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